Hydrostatic pressure Consider a tank of fluid which contains a very thin plate of (neutrally buoyant) material with area A. This situation is shown in Figure below. If the plate is in equilibrium (it does not start to move), then the forces on it must be balanced. On the top side of the plate acts the weight of the fluid, with density ρ, above it in the downward direction. Because it does not accelerate, there must be a force acting upwards on the underside of the plate - this is the effect of the hydrostatic pressure acting on the plate. The weight of the fluid above the plate is: Fw = density volume of fluid accelaration Balancing this is the hydrostatic pressure, p, acting over the area, A, of the underside of the plate depicted in Figure. Balancing the two forces yields: PA = ρgah Cancelling A gives the expression for the hydrostatic pressure as: PA = ρgah, in N (Pascals) Eq. 1 m2 Noting that h is the vertical depth. You can now see the relationship between hydrostatic pressure and gravitational forces. If we evaluated the expression above, we would not find the total pressure acting on the plate. This is because, using the same arguments, there is a column of atmospheric air above the water in the tank and this presses down on the water which, in turn, transmits the additional pressure onto the plate. If we denote the air pressure at the liquid surface PA (again, in N/m2), then the total pressure at a depth h is: P abs = ρgh + P A Eq. 2 This is called the absolute pressure whilst the pressure of Eq. 1 is called the gauge pressure, i.e. the pressure relative to atmospheric pressure.

Pressure modes If a vessel were to be completely empty, containing no molecules whatsoever the pressure would be zero. When you are using this zero as the pressure reference point it is called absolute pressure because there is no lower pressure than the absence of all molecules. There are many applications were measuring pressure is not really dependent on the absolute pressure but the difference between it and the pressure of the atmosphere. When using the atmospheric pressure as the reference point we call the mode gauge pressure. The classic example is a tire, on a typical tire the pressure to be 30 psi (207 kpa) above atmospheric pressure as 0 gauge pressure would mean a flat tire, even though there is technically still atmospheric air pressure in it. The difference between the absolute pressure and gauge pressure value is the variable value of atmospheric pressure: Absolute Pressure = gauge Pressure + atmospheric Pressure And the vacuum pressure can be shown by: Absolute Pressure = gauge Pressure vacuum Pressure Where gauge pressure can be illustrated by: Pressure Measurements: 1- The Manometer P gauge = ρgh It is commonly used to measure small and moderate pressure differences. Consider the manometer shown above that is used to measure the pressure in the tank. Since the gravitational effects of gases are negligible, the pressure anywhere in the tank and at position 1 has the same value. Furthermore, since pressure in a fluid does not vary in the horizontal direction within a fluid, the pressure at point 2 is the

same as the pressure at point 1, P2 = P1. The differential fluid column of height h is in static equilibrium, and it is open to the atmosphere. Then the pressure at point 2 is determined directly from: P 2 = ρgh + P atm where ρ is the density of the fluid in the tube. Note that the cross-sectional area of the tube has no effect on the differential height h, and thus the pressure exerted by the fluid. However, the diameter of the tube should be large enough (more than a few millimeters) to ensure that the surface tension effect and thus the capillary rise is negligible. Example 1: Measuring Pressure with a Manometer. A manometer is used to measure the pressure in a tank. The fluid used has a specific gravity of 0.85, and the manometer column height is 55 cm, as shown in Figure below. If the local atmospheric pressure is 96 kpa, determine the absolute pressure within the tank. Solution: Now with the specific gravity it easy to find the density of the manometer fluid to be: SPg = And the pressure can be calculated by: Fluid density water density 0.85 = ρ f 1000 ρ f = 850 kg m 3 P = ρgh + P atm P = 850 9.81 0.55 + 96000 = 100.586 kpa Where the gauge pressure is: P gauge = ρgh = 850 9.81 0.55 = 4.586 kpa

Case study: What if the manometer has more than one fluid as shown in the flowing figure: Where, the manometer here has three different fluid on three layers. So, the pressure can be calculated easily by: Case study: P at point 1 = ρ 1 gh 1 + ρ 2 gh 2 +ρ 3 gh 3 + P atm Manometers are particularly well-suited to measure pressure drops across a horizontal flow section between two specified points due to the presence of a device such as a valve or heat exchanger or any resistance to flow. This is done by connecting the two legs of the manometer to these two points, as shown in Figure below. The working fluid can be either a gas or a liquid whose density is ρ 1. The density of the manometer fluid is ρ 2, and the differential fluid height is h. A relation for the pressure difference P1 = P2 can be obtained by starting at point 1 with P1, moving along the tube by adding or subtracting the ρgh terms until we reach point 2, and setting the result equal to P2: ρ 1 g(h + a) + P 1 = ρ 1 ga + ρ 2 gh + P 2 ρ 1 gh + ρ 1 ga + P 1 = ρ 1 ga + ρ 2 gh + P 2 P 1 P 2 = ρ 2 gh ρ 1 gh Note that we jumped from point A horizontally to point B and ignored the part underneath since the pressure at both points is the same.

Example 2: The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Figure below. The tank is located on a mountain at an altitude of 1400 m where the atmospheric pressure is 85.6 kpa. Determine the air pressure in the tank if h1 =0.1 m, h2 =0.2 m, and h3=0.35 m. Take the densities of water, oil, and mercury to be 1000 kg/m 3, 850 kg/m 3, and 13,600 kg/m 3, respectively. Solution: now based on the figure: P 1 + ρ 1 gh 1 + ρ 2 gh 2 = P 2 + ρ 3 gh 3 P 1 + 1000 9.81 0.1 + 850 9.81 0.2 = 85600 + 13600 9.81 0.35 P 1 = 131.3 kpa 2- The Barometer and Atmospheric Pressure Atmospheric pressure is measured by a device called a barometer; thus, the atmospheric pressure is often referred to as the barometric pressure.

The pressure at point B is equal to the atmospheric pressure, and the pressure at C can be taken to be zero since there is only mercury vapor above point C and the pressure is very low relative to Patm and can be neglected to an excellent approximation. Writing a force balance in the vertical direction gives. P atm = ρgh where ρ is the density of mercury, g is the local gravitational acceleration, and h is the height of the mercury column above the free surface. Note that the length and the cross-sectional area of the tube have no effect on the height of the fluid column of a barometer. Example 3: Determine the atmospheric pressure at a location where the barometric reading is 740 mm Hg and the gravitational acceleration is g =9.81 m/s2. Assume the temperature of mercury to be 10 C, at which its density is 13,570 kg/m 3. Solution: The barometric reading at a location in height of mercury column is given. The atmospheric pressure is to be determined. P atm = ρgh P atm = 13570 9.81 0.74 = 98.5 kpa Note that density changes with temperature, and thus this effect should be considered in calculations. Example 4: The piston of a vertical piston cylinder device containing a gas has a mass of 60 kg and a cross-sectional area of 0.04 m 2, as shown in Figure below. The local atmospheric pressure is 0.97 bar, and the gravitational acceleration is 9.81 m/s 2. Determine the pressure inside the cylinder. Solution: A gas is contained in a vertical cylinder with a heavy piston. The pressure inside the cylinder and the effect of volume change on pressure are to be determined. To solve it can easily make a force balance on the piston: the atmospheric pressure + the weight of the piston = pressure of the fluid P atm A + w = P A Where, w = mg w = 60 9.81 = 588.6 N, and now:

P = 588.6 0.04 + 97000 = 111.7 kpa If the gas behaves as an ideal gas, the absolute temperature doubles when the volume is doubled at constant pressure. Case Study The pressure balance can be illustrated: Where, γ = ρg, and sinθ = h 2 l 2 P A + γ 1 h 1 = P B + γ 3 h 3 + γ 2 l 2 sinθ

Example 5: The underground storage tank used in a service station contains gasoline filled to the level A. Determine the gage pressure at each of the five identified points. Note that point B is located in the stem, and point C is just below it in the tank. Take gasoline density = 730 kg/m 3. Solution: Since the tube is open-ended, point A is subjected to atmospheric pressure, which has zero-gauge pressure. p A = 0 The pressures at points B and C are the same since they are at the same horizontal level with h = 1 m. p B = p C = ρ gasoline gh = 730 9.81 1 = 7161.3 pa For the same reason, pressure at points D and E is the same. Here, h = 1 m+ 2 m = 3 m. p D = p E = ρ gasoline gh = 730 9.81 3 = 21.48 kpa Example 6: The soaking bin contains ethyl alcohol used for cleaning automobile parts. If h = 7 ft, determine the pressure developed at point A and at the air surface B within the enclosure. Take gea = 49.3 lb/ft 3. Solution: The gauge pressures at points A and B are. p A = γ ea h A = 49.3 (7 2) = 246.5 lb lb = 1.7 = 1.7 psi ft2 in2 p B = γ ea h B = 49.3 (7 6) = 49.3 lb lb = 0.342 = 1.7 psi ft2 in2

Example 7: The funnel is filled with oil and water to the levels shown. Determine the depth of oil h that must be in the funnel so that the water remains at a depth C, and the mercury level made h = 0.8 m. Take oil density = 900 kg/m 3, water density = 1000 kg/m 3, mercury density = 13550 kg/m 3. Solution: the level C to D can be calculated by: And based on the figure: h CD = 0.2 + h + 0.4 0.8 = h 0.2 p A + ρ oil g h + ρ water g 0.4 = p D + ρ Hg g (h 0.2) h = 0.2458 m Note: pressure at the end bottom with height 0.1 m, it is equal for both side so we can eliminate the term, it will be as below: p A + ρ oil g h + ρ water g 0.4 + ρ Hg g 0.1 = p D + ρ Hg g (h 0.2) + ρ Hg g 0.1 So it is same and can be deleted p A + ρ oil g h + ρ water g 0.4 = p D + ρ Hg g (h 0.2)

Example 8: Butyl carbitol, used in the production of plastics, is stored in a tank having a U-tube manometer. If the U-tube is filled with mercury to level E, determine the pressure in the tank at point A and B. Take SHg = 13.55, and Sbc = 0.957. Solution: based on the figure: p A + ρ water g (0.3 + 0.05) = p E + ρ mercury g 0.12 p A + 0.957 1000 9.81 0.35 = 13.55 1000 9.81 0.12 p A = 12.7 kpa Note: the pressure at point E is open to atmosphere so, it is zero. And the pressure at point B: p B + ρ water g (0.25 0.05) = p E + ρ mercury g 0.12 p B + 0.957 1000 9.81 0.2 = 13.55 1000 9.81 0.12 p B = 17.8 kpa Example 9: Water in the reservoir is used to control the water pressure in the pipe at A. If h = 200 mm, determine this pressure when the mercury is at the elevation shown. Take density of mercury = 13 550 kg/m 3. Neglect the diameter of the pipe. Solution: based on the figure: p A + ρ water g 0.25 = p E + ρ water g (0.2 + 0.4 + 0.15) + ρ mercury g 0.1

p A = 18.2 kpa Example 10: A solvent used for plastics manufacturing consists of cyclohexanol in pipe A and ethyl lactate in pipe B that are being transported to a mixing tank. Determine the pressure in pipe A if the pressure in pipe B is 15 psi. The mercury in the manometer is in the position shown, where h = 1 ft. Neglect the diameter of the pipe. Take Sc = 0.953, SHg = 13.55, and Sel = 1.03 Solution: based on the figure: p A + γ cyclohexanol 1.5 + γ mercury 1 = p B + γ ethyl 0.5 p A + 0.953 62.4 1.5 + 13.55 62.4 1 = 2160 + 1.03 62.4 0.5 p A = 1257.4 lb ft 2 p A = 8.73 psi Note: the units of pressure, psi = lb in 2, and the ft = 12 in. Example 11: The two pipes contain hexylene glycol, which causes the level of mercury in the manometer to be at h = 0.3 m. Determine the differential pressure in the pipes, pa - pb. Take ρhgl = 923 kg/m 3, ρhg = 13 550 kg/m 3. Neglect the diameter of the pipes. Solution: based on the figure: p B + γ hgl 0.1 + γ Hg 0.3 = p A + γ hgl 0.1

Test yourself: p A p B = γ Hg 0.3 = 39.9 kpa 1- The pressure in the tank at the closed valve A is 300 kpa. If the differential elevation in the oil level in h = 2.5 m, determine the pressure in the pipe at B. Take oil density = 900 kg/m 3. Answer(329kpa) 2- The two tanks A and B are connected using a manometer. If waste oil is poured into tank A to a depth of h = 0.6 m, determine the pressure of the entrapped air in tank B. Air is also trapped in line CD as shown. Take oil density = 900 kg/m 3, water density= 1000 kg/m 3. Answer (15.1 kpa) 3- Air is pumped into the water tank at A such that the pressure gage reads 20 psi. Determine the pressure at point B at the bottom of the ammonia tank. Take ammonia density = 1.75 slug/ft 3. Answer (22.6psi)

4- The Morgan Company manufactures a micromanometer that works on the principles shown. Here there are two reservoirs filled with kerosene, each having a crosssectional area of 300 mm 2. The connecting tube has a crosssectional area of 15 mm 2 and contains mercury. Determine h if the pressure difference pa - pb = 40 Pa. What would h be if water were substituted for mercury? Mercury density= 13550 kg/m 3, Ke density= 814 kg/m 3. Answer (h=18mm) 5- Determine the difference in pressure pb pa between the centers A and B of the pipes, which are filled with water. The mercury in the inclined-tube manometer has the level shown SHg = 13.55. Answer(18.5kpa)